Electromagnetism - Faraday’s law

A circular wire loop of radius R rotates with an angular speed ω in a uniform magnetic field B, as shown in the figure. If the emf ɛ induced in the loop is ɛ₀ sin ωt, then the angular speed of the loop is

A. ɛ₀ R/B
B. 2πɛ₀/R
C. ɛ₀/BπR²
D. ɛ₀²/BR²
E. tan⁻¹(ɛ₀/Bc)

(GR9677 #46)

Solution:

Faraday's Law: ɛ = − dΦ/dt
with  Φ = NBA  

Given:
N = 1
B uniform (constant)
ɛ =  ɛ₀ sin ωt

→ ɛ = − B dA/dt =  ɛ₀ sin ωt

− B dA =  ɛ₀ sin ωt dt 

 B ∫dA =  ɛ₀ ∫ sin ωt dt 

− BπR² = − ^ɛ₀/_ω  cos ωt

ω = ɛ₀ cos ωt / BπR²

at t = 0,
ω = ɛ₀ / BπR²

Answer: C

Alternative solution:
A = πR²  cos ωt
ɛ = − dΦ/dt = − B dA/dt 
ɛ = − BπR²  d(cos ωt)/dt 
ɛ₀ sin ωt = BπR²ω  sin ωt 
ɛ₀= BπR²ω  

ω = ɛ₀ / (BπR²)